\section{Test A3}
\def\firstcircle{(0,0) circle (1)}
\def\secondcircle{(1,0) circle (1)}
\def\thirdcircle{(0.5,-0.866025) circle (1)}
%Aufgabe a
\subsection{}
\begin{tikzpicture}[fill=gray]
% A^B
\scope
\clip \secondcircle;
\fill \firstcircle;
\endscope
% A^C
\scope
\clip \thirdcircle;
\fill \firstcircle;
\endscope
% outline
\draw \firstcircle (0,1)  node [text=black,above] {$A$}
      \secondcircle (1,1)  node [text=black,above] {$B$}
      \thirdcircle (0.5,-1.866025)  node [text=black,below] {$C$}
      (-2,-3) rectangle (3,2) node [text=black,above] {$U$};
\end{tikzpicture}

\begin{displaymath}
\begin{center}
Es gilt zu beweisen: $A\cap{}(B\cup{}C) = (A\cap{}B)\cup{}(A\cap{}C)$
\end{center}
\begin{array}{ll}
B\cup{}C & = \{x\in{}U|(x\in{}B)\vee{}(x\in{}C)\} \\
A\cap{}(B\cup{}C) & = \{x\in{}U|(x\in{}A)\wedge{}((x\in{}B)\vee{}(x\in{}C))\} \\
A\cap{}B & = \{x\in{}U|(x\in{}A)\wedge{}(x\in{}B)\} \\
A\cap{}C & = \{x\in{}U|(x\in{}A)\wedge{}(x\in{}C)\} \\
(A\cap{}B)\cup{}(A\cap{}C) & = \{x\in{}U|((x\in{}A)\wedge{}(x\in{}B))\vee{}((x\in{}A)\wedge{}(x\in{}C))\} \\
\end{array}
\end{displaymath}
\begin{displaymath}
\begin{array}{|c|c|c||c|c|c|c|c|}
	\hline
   A
 & B
 & C
 & B\vee{}C
 & A\wedge{}(B\vee{}C)
 & A\wedge{}B
 & A\wedge{}C
 & (A\wedge{}B)\vee{}(A\wedge{}C) \\
 \hline \hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
\end{array}
\end{displaymath}
Da alle Aussagen in $A\wedge{}(B\vee{}C)$ und $(A\wedge{}B)\vee{}(A\wedge{}C)$ immer zusammen WAHR oder FALSCH sind, gilt $A\cap{}(B\cup{}C) = (A\cap{}B)\cup{}(A\cap{}C)$

%Aufgabe b
\subsection{}
\begin{tikzpicture}[fill=gray]
%A
\scope
\clip (-2,-3) rectangle (3,2);
\fill \firstcircle;
\endscope
% A^B
\scope
\clip \secondcircle;
\fill \firstcircle;
\endscope
% A^C
\scope
\clip \thirdcircle;
\fill \firstcircle;
\endscope
% B^C
\scope
\clip \thirdcircle;
\fill \secondcircle;
\endscope
% outline
\draw \firstcircle (0,1)  node [text=black,above] {$A$}
      \secondcircle (1,1)  node [text=black,above] {$B$}
      \thirdcircle (0.5,-1.866025)  node [text=black,below] {$C$}
      (-2,-3) rectangle (3,2) node [text=black,above] {$U$};
\end{tikzpicture}

\begin{displaymath}
\begin{center}
Es gilt zu beweisen: $A\cup{}(B\cap{}C) = (A\cup{}B)\cap{}(A\cap{}C)$
\end{center}
\begin{array}{ll}
B\cap{}C & = \{x\in{}U|(x\in{}B)\wedge{}(x\in{}C)\} \\
A\cup{}(B\cap{}C) & = \{x\in{}U|(x\in{}A)\vee{}((x\in{}B)\wedge{}(x\in{}C))\} \\
A\cup{}B & = \{x\in{}U|(x\in{}A)\vee{}(x\in{}B)\} \\
A\cap{}C & = \{x\in{}U|(x\in{}A)\wedge{}(x\in{}C)\} \\
(A\cup{}B)\cap{}(A\cap{}C) & = \{x\in{}U|((x\in{}A)\vee{}(x\in{}B))\wedge{}((x\in{}A)\wedge{}(x\in{}C))\} \\
\end{array}
\end{displaymath}
\begin{displaymath}
\begin{array}{|c|c|c||c|c|c|c|c|}
	\hline
   A
 & B
 & C
 & B\wedge{}C
 & A\vee{}(B\wedge{}C)
 & A\vee{}B
 & A\wedge{}C
 & (A\vee{}B)\wedge{}(A\wedge{}C) \\
 \hline \hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
\hline
0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
\end{array}
\end{displaymath}

\begin{displaymath}
\begin{array}{ll}
A\cup{}C & = \{x\in{}U|(x\in{}A)\vee{}(x\in{}C)\} \\
(A\cup{}B)\cap{}(A\cup{}C) & = \{x\in{}U|((x\in{}A)\vee{}(x\in{}B))\wedge{}((x\in{}A)\vee{}(x\in{}C))\} \\
\end{array}
\end{displaymath}
\begin{displaymath}
\begin{array}{|c|c|c||c|c|c|}
	\hline
   A
 & B
 & C
 & A\vee{}B
 & A\vee{}C
 & (A\vee{}B)\wedge{}(A\vee{}C) \\
 \hline \hline
1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 1 & 1 & 1 \\
\hline
0 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\hline
\end{array}
\end{displaymath}
Da die Aussagen in $A\vee{}(B\wedge{}C)$ und $(A\vee{}B)\wedge{}(A\wedge{}C)$ unterschiedlich WAHR oder FALSCH sind, gilt  $A\cup{}(B\cap{}C) \neq{} (A\cup{}B)\cap{}(A\cap{}C)$. Stattdessen gilt aber $A\cup{}(B\cap{}C) = (A\cup{}B)\cap{}(A\cup{}C)$, da alle Aussagen in $A\vee{}(B\wedge{}C)$ und $(A\vee{}B)\wedge{}(A\vee{}C)$ immer zusammen WAHR oder FALSCH sind.

%Aufgabe c.
\subsection{}
\begin{tikzpicture}[fill=gray]
% U
\scope
\fill (-2,-2) rectangle (3,2);
\endscope
% A
\scope
\fill[white] \firstcircle;
\endscope
% B
\scope
\fill[white] \secondcircle;
\endscope
% outline
\draw \firstcircle (0,1)  node [text=black,above] {$A$}
      \secondcircle (1) (1,1)  node [text=black,above] {$B$}
      (-2,-2) rectangle (3,2) node [text=black,above] {$H$};
\end{tikzpicture}

\begin{displaymath}
\begin{center}
Es gilt zu beweisen: $(A\cup{})^{C} = A^{C}\cap{}B^{C}$
\end{center}
\begin{array}{ll}
B\cup{}C & = \{x\in{}U|(x\in{}B)\vee{}(x\in{}C)\} \\
A\cap{}(B\cup{}C) & = \{x\in{}U|(x\in{}A)\wedge{}((x\in{}B)\vee{}(x\in{}C))\} \\
A\cap{}B & = \{x\in{}U|(x\in{}A)\wedge{}(x\in{}B)\} \\
A\cap{}C & = \{x\in{}U|(x\in{}A)\wedge{}(x\in{}C)\} \\
(A\cap{}B)\cup{}(A\cap{}C) & = \{x\in{}U|((x\in{}A)\wedge{}(x\in{}B))\vee{}((x\in{}A)\wedge{}(x\in{}C))\} \\
\end{array}
\end{displaymath}
\begin{displaymath}
\begin{array}{|c|c|c||c|c|c|c|c|}
	\hline
   A
 & B
 & C
 & B\vee{}C
 & A\wedge{}(B\vee{}C)
 & A\wedge{}B
 & A\wedge{}C
 & (A\wedge{}B)\vee{}(A\wedge{}C) \\
 \hline \hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
\end{array}
\end{displaymath}
Da alle Aussagen in $A\wedge{}(B\vee{}C)$ und $(A\wedge{}B)\vee{}(A\wedge{}C)$ immer zusammen WAHR oder FALSCH sind, gilt $A\cap{}(B\cup{}C) = (A\cap{}B)\cup{}(A\cap{}C)$